calculate mean using while and iteration?

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Madan Kumar
Madan Kumar am 30 Apr. 2019
Kommentiert: Madan Kumar am 3 Mai 2019
Hi,
I have data of ~3500x2. I want to calulate mean of second column for a particular condition in first column using 'while'.
let data (a, b) be like
0.5 1.8
0.6 1.5
0.9 1.8
1.0 1.5
1.1 1.4
1.2 1.4
1.5 1.6
1.8 1.2
2.1 1.2
2.3 1.3
2.4 1.5
2.6 1.8
2.9 2.0
3.0 3.0
3.12 3.2
3.15 1.9
3.16 1.7
3.18 2.2
I need to calculate mean of b, if a> 0.5 and a<1.5. Then increase 'a' by 1 and calculate mean of b (i.e for a > 1.5 and a<2.5) and so on. It may be a silly question but I am stuck with it. My code is
del=0.5;
k=1;
a(k)=1;
while(a(k) >(a(k)-del) && a(k)< (a(k)+del))
xn(k)=mean(b(k));
k= k+1;
a(k)=a(k)+1;
end
but it shows error Index exceeds array bounds.
Error in untitled (line 12)
a(k)=a(k)+1;
Thank you for your help.
  2 Kommentare
David Wilson
David Wilson am 30 Apr. 2019
Bearbeitet: David Wilson am 30 Apr. 2019
My code below is a bit ugly, but I think it does what you want:
cutoff = [0.5 1.5]; % band of interest
maxA = ceil(max(a))+0.5;
bmean = [];
for i=1:maxA
idx = find(a>cutoff(1) & a<cutoff(2));
bmean(i) = mean(b(idx));
cutoff = cutoff+1;
end
The means of column "b" are in variable bmean.
I note that you specified strict < as opposed to <= which may, or may not be what you really want.
Note that column a need not be sorted in increasing order.
Madan Kumar
Madan Kumar am 30 Apr. 2019
Thank you...

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Akzeptierte Antwort

Rik
Rik am 30 Apr. 2019
So you want to calculate these values?
xn(1)=mean(b(a>0.5 & a<1.5));
xn(2)=mean(b(a>1.5 & a<2.5));
xn(3)=mean(b(a>2.5 & a<3.5));
etc?
You don't need a while loop for that:
a=[0.5 0.6 0.9 1.0 1.1 1.2 1.5 1.8 2.1 2.3 2.4 2.6 2.9 3.0 3.12 3.15 3.16 3.18];
b=[1.8 1.5 1.8 1.5 1.4 1.4 1.6 1.2 1.2 1.3 1.5 1.8 2.0 3.0 3.2 1.9 1.7 2.2];
del=0.5;
xn=zeros(1,ceil(max(a-del)));
for k=1:size(xn,2)
xn(k)=mean(b(a>(k-del) & a<(k+del)));
end
  4 Kommentare
Stephen23
Stephen23 am 30 Apr. 2019
Bearbeitet: Stephen23 am 30 Apr. 2019
"Suppose, I need xn(0.5)=... "
You can't. Indices must be whole integers greater than zero.
Either change del to 0.25 and leave it at that, or write a function which lets you have any input values that you desire. But you certainly cannot have indexing with non-integer values.
Madan Kumar
Madan Kumar am 3 Mai 2019
Ok...thank you...

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Weitere Antworten (1)

KSSV
KSSV am 30 Apr. 2019
Why loop? YOu can use inbuilt in mean. Let a,b be your columns.
idx = a>0.5 & a<1.5 ;
mean(b(idx))

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